Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct Node {
  int v;
  Node *x;
} * e, *lnk[100000 + 9], E[2 * 100000 + 9];
int n, m, q, u, v, w, a;
int d[100000 + 9], f[100000 + 9][19];
vector<int> g[100000 + 9], h[100000 + 9][19];
vector<int> ans;
void add(int u, int v) { e->v = v, e->x = lnk[u], lnk[u] = e++; }
vector<int> mrg(const vector<int> &a, const vector<int> &b, int l = 10) {
  vector<int> c;
  int x = 0, y = 0;
  while (c.size() < l) {
    if (x < a.size() && y < b.size())
      c.push_back(a[x] < b[y] ? a[x++] : b[y++]);
    else if (x < a.size())
      c.push_back(a[x++]);
    else if (y < b.size())
      c.push_back(b[y++]);
    else
      break;
  }
  return c;
}
void dfs(int fat, int u) {
  d[u] = d[fat] + 1;
  f[u][0] = fat;
  h[u][0] = g[fat];
  for (int i = 1; i < 19; ++i) {
    f[u][i] = f[f[u][i - 1]][i - 1];
    h[u][i] = mrg(h[u][i - 1], h[f[u][i - 1]][i - 1]);
  }
  for (Node *p = lnk[u]; p; p = p->x)
    if (p->v != fat) dfs(u, p->v);
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  e = E;
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &u, &v);
    add(u, v);
    add(v, u);
  }
  for (int i = 1; i <= m; ++i) {
    scanf("%d", &w);
    g[w].push_back(i);
  }
  for (int i = 1; i <= n; ++i) {
    sort(g[i].begin(), g[i].end());
    if (g[i].size() > 10) g[i].resize(10);
  }
  dfs(0, 1);
  while (q--) {
    ans.clear();
    scanf("%d%d%d", &u, &v, &a);
    if (d[u] < d[v]) swap(u, v);
    ans = g[u];
    if (ans.size() > a) ans.resize(a);
    for (int k = 0; k < 19; ++k)
      if (d[u] - d[v] >> k & 1) {
        ans = mrg(ans, h[u][k], a);
        u = f[u][k];
      }
    if (u != v) {
      ans = mrg(ans, g[v], a);
      for (int k = 18; k >= 0; --k)
        if (f[u][k] != f[v][k]) {
          ans = mrg(ans, h[u][k], a);
          u = f[u][k];
          ans = mrg(ans, h[v][k], a);
          v = f[v][k];
        }
      ans = mrg(ans, h[u][0], a);
      u = f[u][0];
    }
    printf("%d", ans.size());
    for (auto i : ans) printf(" %d", i);
    putchar('\n');
  }
  return 0;
}